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I Ladder slipping against a smooth wall

  1. May 12, 2016 #1
    Consider a ladder slanted like \ of length ##l## slipping against a smooth wall and on a smooth floor.

    I come to the contradiction that there must be a deceleration in the x direction but there is no force opposing the velocity of the ladder.

    Its free-body diagram contains a rightward normal force at the top end and a downward weight at the center of gravity.


    Let the coordinates of the top and bottom ends be ##(0, y)## and ##(x, 0)## respectively. We have ##x^2+y^2=l^2##.

    Let the distance moved by the top and bottom ends in time ##\Delta t## be ##\Delta y## and ##\Delta x## respectively. We have ##(x+\Delta x)^2+(y-\Delta y)^2=l^2##.

    Ignoring higher powers of ##\Delta x## and ##\Delta y## and also using the first equation, we get ##\frac{\Delta y}{\Delta x}=\frac{x}{y}##. Dividing both the numerator and denominator by ##\Delta t##, we get ##\frac{v_y}{v_x}=\frac{x}{y}##. This means that ##v_x>0## shortly after the slipping begins, and when the ladder just hits the floor, ##y=0## and thus ##v_x=0##. In other words, there is a deceleration in the x direction. But then there is no leftward force in the free-body diagram.

    What's wrong?
     
    Last edited: May 12, 2016
  2. jcsd
  3. May 12, 2016 #2

    BvU

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    Are you saying you can't place the ladder against the wall without having ##v_x>0## ?
    How does ##y = 0## imply ##v_x = 0## ?
     
  4. May 12, 2016 #3
    ##v_x>0## shortly after the slipping begins when ##v_y>0##.

    ##v_x=\frac{y}{x}v_y## so ##v_x=0## when ##y=0##.
     
  5. May 12, 2016 #4

    BvU

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    Haven't found it yet. Center of mass moves over the dotted circle. Momentum to the right is traded in for angular momentum that makes ##v_x=0## at the end.

    upload_2016-5-13_2-11-9.png
     
  6. May 13, 2016 #5
    If the upward force acting at the bottom end is ##mg##, then there is no net force in the y direction. Consequently, the center of mass would not fall down, producing a contradiction.

    Is the upward force zero?
     
  7. May 13, 2016 #6

    BvU

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    I pictured the situation at the moment the restraining peg at the foot of the ladder is removed. It kept the thing in place by compensating the normal force that the wall exerts.
    initially the ladder is stationary.

    I'm off for the weekend. @haruspex , can you help us ?
     
  8. May 13, 2016 #7

    haruspex

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    The ladder will not stay in contact with the wall.
    Challenge: show that it will leave the wall when the angle to the vertical is arcos(2/3).

    It is not even immediately obvious that it will stay in contact with the floor! At first, it must have substantial rotational acceleration in comparison to its vertical acceleration, so it is feasible that later on it will have so much angular velocity in relation to its vertical velocity that it could become airborne (but it doesn't).
     
  9. May 13, 2016 #8
    What if the initial angle to the vertical is already bigger than arcos(2/3)? Does it mean that it would immediately lose contact with the wall once released?
     
    Last edited: May 13, 2016
  10. May 13, 2016 #9

    haruspex

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    No, it needs to build up some horizontal speed first.
     
  11. May 13, 2016 #10
    Then it seems like the angle at which the ladder leaves the wall depends on the initial angle. So the initial angle has to be given in the challenge you posted, right?
     
  12. May 13, 2016 #11

    haruspex

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    The challenge I posted assumed the context of the original question in the thread, which I took to be almost vertical. But now I look back and don't see that stated.
     
  13. Jul 14, 2016 #12
    I figured the trick to the challenge is to imagine the center of mass of the ladder as a point mass sliding from the top of a frictionless sphere. The angle the position vector of this mass to the vertical is ##\cos^{-1}(2/3)## when the mass leaves the sphere. Then the angle the ladder makes with the vertical should be 90##^\circ-\cos^{-1}(2/3)##.

    How do you show that the ladder never become airborne? The point mass leaving the sphere will continue in a parabolic path. What about the ladder? What's its path and tilt after leaving the wall? It is not possible for the center of mass to move in a parabola while maintaining the same tilt (because the lower end of the ladder cannot penetrate the floor). The tilt must become gentler as the ladder falls. So must there be an upward force at the point of contact with the floor to create an anticlockwise moment to reduce the tilt? It seems the ladder already has an angular momentum, which can reduce its tilt. How can we solve for its path and tilt?
     
    Last edited: Jul 14, 2016
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